#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define endl '\n'
const int N = 2e5 + 5;

ll a, b;
int c[15], num;
int d;
ll res[10];
void init(ll n) {
  num = 0;
  while (n) {
    c[++num] = n % 10;
    n /= 10;
  }
  if (!num) c[++num] = 0;
}
ll dp[2][2][15], cnt[2][2][15];
ll dfs(bool zero, bool limit, int s) {
  if (s == 0) {
    cnt[zero][limit][s] = 1;
    return dp[zero][limit][s] = 0;
  }
  if (dp[zero][limit][s] != -1) return dp[zero][limit][s];
  dp[zero][limit][s] = 0;
  cnt[zero][limit][s] = 0;
  int imx = limit ? c[s] : 9;
  rep(i, 0, imx) {
    int czero = zero && i == 0;
    int climit = limit && i == imx;
    dp[zero][limit][s] += dfs(czero, climit, s - 1);
    if (i == d && !czero) dp[zero][limit][s] += cnt[czero][climit][s - 1];
    cnt[zero][limit][s] += cnt[czero][climit][s - 1];
  }
  return dp[zero][limit][s];
}
void solve() {
  init(b);
  rep(i, 0, 9) {
    d = i;
    memset(dp, -1, sizeof(dp));
    res[i] = dfs(1, 1, num);
  }
  init(a - 1);
  rep(i, 0, 9) {
    d = i;
    memset(dp, -1, sizeof(dp));
    res[i] -= dfs(1, 1, num);
  }
  rep(i, 0, 9) cout << res[i] << ' ';
}

int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> a >> b;
  solve();
  return 0;
}